The proof is by contradiction.
Step 1. Aim for this step is to prove that, assuming equality holds in Cheeger's inequality, we obtain the existence of a function \(f\in D(\Delta )\)  such that \(-\Delta f=\lambda _0 f\)  and
\(|\nabla f|=\sqrt{\lambda _0}\,|f|,\,\,\,\text{$\mathfrak {m}$-a.e. .}\) 
First of all, under our assumptions thm:compembedding implies that \(\lambda _0\)  (and thus also \(h(X)\) ) is positive and that there exists \(f\in D(\Delta ) \subset W^{1,2}(X,\mathsf {d},\mathfrak {m})\)  with \(-\Delta f= \lambda _{0}f\) , and \(\Vert f\Vert _2=1\) .
From Lemma REF  we know that
\(|D (f)^{2}|(X)\le 2 \int _{X} |f|\, | \nabla f| \, \mathsf {d}\mathfrak {m}, \) 
and we can apply the Cauchy-Schwarz inequality to infer
\(2\left( \int _X |\nabla f|^2 \, \mathsf {d}\mathfrak {m}\right)^{\frac{1}{2}}\left( \int _X |f|^2 \, \mathsf {d}\mathfrak {m}\right)^{\frac{1}{2}}\ge |D(f)^{2}|(X)\) 
Using the coarea formula (REF ) and recalling that \(\Vert f\Vert _2=1\) , we obtain
\(\begin{aligned}& 2\sqrt{\lambda _0}=2\bigg ( \int _X | \nabla f|^2 \, \mathsf {d}\mathfrak {m}\bigg )^{\frac{1}{2}}\overset{(\ref {eq:CSProof})}{\ge } |D(f)^{2}|(X) \overset{ (\ref {eq:coareaBV})}{=} \int _{0}^{\infty } \mathrm {Per}(\lbrace (f)^{2}>t\rbrace ) \, \mathsf {d}t \\&\ge h(X)\int _{0}^{\infty } \mathfrak {m}(\lbrace (f)^{2}>t\rbrace ) \, \mathsf {d}t= h(X)\int _{X} f^{2} \, \mathsf {d}\mathfrak {m}=h(X)=2\sqrt{\lambda _0},\end{aligned}\) 
where the last identity comes from the assumption that equality is achieved in Cheeger's inequality. It follows that all the inequalities in (REF ) are actually equalities. In particular
\(\bigg ( \int _X |\nabla f|^2 \, \mathsf {d}\mathfrak {m}\bigg )^{\frac{1}{2}}=\int _X |\nabla f||f|\,\mathsf {d}\mathfrak {m}\, .\) 
By the equality case (REF ) in the Cauchy-Schwartz inequality we can infer that \(|\nabla f|=C|f|\)  \(\mathfrak {m}\) -a.e., for some constant \(C> 0\) . Since
\(\lambda _0=\int _X |\nabla f|^2\,\mathsf {d}\mathfrak {m}=\int _X C^2f^2\,\mathsf {d}\mathfrak {m}=C^2\) 
the claim (REF ) follows.
Step 2.
Since \(f\in L^{\infty }(X,\mathfrak {m})\)  and \(-\Delta f=\lambda _0f\)  we have that \(\Delta f\in L^{\infty }(X,\mathfrak {m})\)  and also (up to a suitable choice of \(\mathfrak {m}\) -a.e. representative) \(f\in \mathsf {Lip}_b(X)\)  by the \(L^{\infty }-\mathsf {Lip}\)  regularization of the heat semigroup (REF ).
We are now in position to apply [1]}. Thus, for any \(T>0\)  we can find a test plan \(\Pi \in \mathcal {P}(C([0,T],X))\)  such that \(({\rm e}_0)_{\sharp }\Pi =\mathfrak {m}\)  and
\(f(\gamma (t))-f(\gamma (s))=\int _s^t|\nabla f|^2(\gamma (r))\mathsf {d}r=\int _s^t\lambda _0f^2(\gamma (r))\mathsf {d}r,\) 
for any \(0\le s\le t\le T\)  and for \(\Pi \) -almost every \(\gamma \) . Observe that, fixing any such curve \(\gamma \)  and setting \(F(t):=f(\gamma (t))\) , \(F\)  is smooth and it verifies the ODE
\(F^{\prime }(t)=\lambda _0F^2(t)\) 
in the classical sense. Indeed it is absolutely continuous, it solves the ODE in the almost everywhere sense and the derivative itself is continuous, allowing to bootstrap the regularity.
Observe that if \(F(0)=0\) , then \(F(t)=0\)  for every \(t\in [0,T]\) . Otherwise, if \(F(0)>0\)  there is no bounded solution of (REF ) up to time \(1/(\lambda _0F(0))\) .
We claim that \(f\)  vanishes identically, contradicting the assumption \(\int _X f^2\, \mathsf {d}\mathfrak {m}=1\) . If this is not the case we can find \(\epsilon >0\)  and a set of positive measure \(E\subset X\)  such that for any \(x\in E\)  it holds \(f(x)\lambda _0>\epsilon \) . Then we apply the construction above with \(T=1/\epsilon \)  finding \(\Pi \in \mathcal {P}(C([0,T],X))\)  verifying (REF ) and such that, for a set of positive \(\Pi \) -measure of curves \(\gamma \) , it holds \(f(\gamma (0))\lambda _0>\epsilon \) . Recalling what we pointed out above concerning bounded solutions of (REF ), we obtain a contradiction, since \(f\)  is bounded.
